zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

tan^2(x)+tan(x)+cot(x)+cot^2(x)=4

解答

tan^{2} (x) + tan (x) + cot (x) + cot^{2} (x) = 4

解答

x = \frac{π}{4} + πn, x = 1.93566 \dots + πn, x = 2.77672 \dots + πn

+1

度数

x = 4 5^{\circ} + 18 0^{\circ} n, x = 110.90515 \dots^{\circ} + 18 0^{\circ} n, x = 159.09484 \dots^{\circ} + 18 0^{\circ} n

求解步骤

tan^{2} (x) + tan (x) + cot (x) + cot^{2} (x) = 4

两边减去

4

tan^{2} (x) + tan (x) + cot (x) + cot^{2} (x) - 4 = 0

使用三角恒等式改写

- 4 + cot (x) + cot^{2} (x) + tan (x) + tan^{2} (x)

使用基本三角恒等式:

tan (x) = \frac{1}{cot ( x )}

= - 4 + cot (x) + cot^{2} (x) + \frac{1}{cot ( x )} + (\frac{1}{cot ( x )})^{2}

(\frac{1}{cot ( x )})^{2} = \frac{1}{cot ^{2} ( x )}

(\frac{1}{cot ( x )})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{cot ^{2} ( x )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{cot ^{2} ( x )}

= - 4 + cot (x) + cot^{2} (x) + \frac{1}{cot ( x )} + \frac{1}{cot ^{2} ( x )}

- 4 + cot (x) + cot^{2} (x) + \frac{1}{cot ^{2} ( x )} + \frac{1}{cot ( x )} = 0

用替代法求解

- 4 + cot (x) + cot^{2} (x) + \frac{1}{cot ^{2} ( x )} + \frac{1}{cot ( x )} = 0

令

: cot (x) = u

- 4 + u + u^{2} + \frac{1}{u ^{2}} + \frac{1}{u} = 0

- 4 + u + u^{2} + \frac{1}{u ^{2}} + \frac{1}{u} = 0 : u = 1, u = \frac{- 3 + 5}{2}, u = \frac{- 3 - 5}{2}

- 4 + u + u^{2} + \frac{1}{u ^{2}} + \frac{1}{u} = 0

乘以最小公倍数

- 4 + u + u^{2} + \frac{1}{u ^{2}} + \frac{1}{u} = 0

找到

u^{2}, u

的最小公倍数:

u^{2}

u^{2}, u

最小公倍数 (LCM)

计算出由出现在

u^{2}

或

u

中的因子组成的表达式

= u^{2}

乘以最小公倍数=

u^{2}

- 4 u^{2} + u u^{2} + u^{2} u^{2} + \frac{1}{u ^{2}} u^{2} + \frac{1}{u} u^{2} = 0 \cdot u^{2}

化简

- 4 u^{2} + u u^{2} + u^{2} u^{2} + \frac{1}{u ^{2}} u^{2} + \frac{1}{u} u^{2} = 0 \cdot u^{2}

化简

u u^{2} : u^{3}

u u^{2}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

u u^{2} = u^{1 + 2}

= u^{1 + 2}

数字相加:

1 + 2 = 3

= u^{3}

化简

u^{2} u^{2} : u^{4}

u^{2} u^{2}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u^{2} = u^{2 + 2}

= u^{2 + 2}

数字相加:

2 + 2 = 4

= u^{4}

化简

\frac{1}{u ^{2}} u^{2} : 1

\frac{1}{u ^{2}} u^{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot u ^{2}}{u ^{2}}

约分:

u^{2}

= 1

化简

\frac{1}{u} u^{2} : u

\frac{1}{u} u^{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot u ^{2}}{u}

乘以:

1 \cdot u^{2} = u^{2}

= \frac{u ^{2}}{u}

约分:

u

= u

化简

0 \cdot u^{2} : 0

0 \cdot u^{2}

使用法则

0 \cdot a = 0

= 0

- 4 u^{2} + u^{3} + u^{4} + 1 + u = 0

- 4 u^{2} + u^{3} + u^{4} + 1 + u = 0

- 4 u^{2} + u^{3} + u^{4} + 1 + u = 0

解

- 4 u^{2} + u^{3} + u^{4} + 1 + u = 0 : u = 1, u = \frac{- 3 + 5}{2}, u = \frac{- 3 - 5}{2}

- 4 u^{2} + u^{3} + u^{4} + 1 + u = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{4} + u^{3} - 4 u^{2} + u + 1 = 0

因式分解

u^{4} + u^{3} - 4 u^{2} + u + 1 : (u - 1)^{2} (u^{2} + 3 u + 1)

u^{4} + u^{3} - 4 u^{2} + u + 1

使用有理根定理

a_{0} = 1, a_{n} = 1

a_{0}

的除数

: 1, a_{n}

的除数

: 1

因此，检验以下有理数:

\pm \frac{1}{1}

\frac{1}{1}

是表达式的根，所以因式分解

u - 1

= (u - 1) \frac{u ^{4} + u ^{3} - 4 u ^{2} + u + 1}{u - 1}

\frac{u ^{4} + u ^{3} - 4 u ^{2} + u + 1}{u - 1} = u^{3} + 2 u^{2} - 2 u - 1

\frac{u ^{4} + u ^{3} - 4 u ^{2} + u + 1}{u - 1}

对

\frac{u ^{4} + u ^{3} - 4 u ^{2} + u + 1}{u - 1}

做除法:

\frac{u ^{4} + u ^{3} - 4 u ^{2} + u + 1}{u - 1} = u^{3} + \frac{2 u ^{3} - 4 u ^{2} + u + 1}{u - 1}

将分子

u^{4} + u^{3} - 4 u^{2} + u + 1

与除数

u - 1

的首项系数相除：

\frac{u ^{4}}{u} = u^{3}

商 = u^{3}

将

u - 1

乘以

u^{3} : u^{4} - u^{3}

将

u^{4} + u^{3} - 4 u^{2} + u + 1

减去

u^{4} - u^{3}

得到新的余数

余数 = 2 u^{3} - 4 u^{2} + u + 1

因此

\frac{u ^{4} + u ^{3} - 4 u ^{2} + u + 1}{u - 1} = u^{3} + \frac{2 u ^{3} - 4 u ^{2} + u + 1}{u - 1}

= u^{3} + \frac{2 u ^{3} - 4 u ^{2} + u + 1}{u - 1}

对

\frac{2 u ^{3} - 4 u ^{2} + u + 1}{u - 1}

做除法:

\frac{2 u ^{3} - 4 u ^{2} + u + 1}{u - 1} = 2 u^{2} + \frac{- 2 u ^{2} + u + 1}{u - 1}

将分子

2 u^{3} - 4 u^{2} + u + 1

与除数

u - 1

的首项系数相除：

\frac{2 u ^{3}}{u} = 2 u^{2}

商 = 2 u^{2}

将

u - 1

乘以

2 u^{2} : 2 u^{3} - 2 u^{2}

将

2 u^{3} - 4 u^{2} + u + 1

减去

2 u^{3} - 2 u^{2}

得到新的余数

余数 = - 2 u^{2} + u + 1

因此

\frac{2 u ^{3} - 4 u ^{2} + u + 1}{u - 1} = 2 u^{2} + \frac{- 2 u ^{2} + u + 1}{u - 1}

= u^{3} + 2 u^{2} + \frac{- 2 u ^{2} + u + 1}{u - 1}

对

\frac{- 2 u ^{2} + u + 1}{u - 1}

做除法:

\frac{- 2 u ^{2} + u + 1}{u - 1} = - 2 u + \frac{- u + 1}{u - 1}

将分子

- 2 u^{2} + u + 1

与除数

u - 1

的首项系数相除：

\frac{- 2 u ^{2}}{u} = - 2 u

商 = - 2 u

将

u - 1

乘以

- 2 u : - 2 u^{2} + 2 u

将

- 2 u^{2} + u + 1

减去

- 2 u^{2} + 2 u

得到新的余数

余数 = - u + 1

因此

\frac{- 2 u ^{2} + u + 1}{u - 1} = - 2 u + \frac{- u + 1}{u - 1}

= u^{3} + 2 u^{2} - 2 u + \frac{- u + 1}{u - 1}

对

\frac{- u + 1}{u - 1}

做除法:

\frac{- u + 1}{u - 1} = - 1

将分子

- u + 1

与除数

u - 1

的首项系数相除：

\frac{- u}{u} = - 1

商 = - 1

将

u - 1

乘以

- 1 : - u + 1

将

- u + 1

减去

- u + 1

得到新的余数

余数 = 0

因此

\frac{- u + 1}{u - 1} = - 1

= u^{3} + 2 u^{2} - 2 u - 1

= u^{3} + 2 u^{2} - 2 u - 1

分解

u^{3} + 2 u^{2} - 2 u - 1 : (u - 1) (u^{2} + 3 u + 1)

u^{3} + 2 u^{2} - 2 u - 1

使用有理根定理

a_{0} = 1, a_{n} = 1

a_{0}

的除数

: 1, a_{n}

的除数

: 1

因此，检验以下有理数:

\pm \frac{1}{1}

\frac{1}{1}

是表达式的根，所以因式分解

u - 1

= (u - 1) \frac{u ^{3} + 2 u ^{2} - 2 u - 1}{u - 1}

\frac{u ^{3} + 2 u ^{2} - 2 u - 1}{u - 1} = u^{2} + 3 u + 1

\frac{u ^{3} + 2 u ^{2} - 2 u - 1}{u - 1}

对

\frac{u ^{3} + 2 u ^{2} - 2 u - 1}{u - 1}

做除法:

\frac{u ^{3} + 2 u ^{2} - 2 u - 1}{u - 1} = u^{2} + \frac{3 u ^{2} - 2 u - 1}{u - 1}

将分子

u^{3} + 2 u^{2} - 2 u - 1

与除数

u - 1

的首项系数相除：

\frac{u ^{3}}{u} = u^{2}

商 = u^{2}

将

u - 1

乘以

u^{2} : u^{3} - u^{2}

将

u^{3} + 2 u^{2} - 2 u - 1

减去

u^{3} - u^{2}

得到新的余数

余数 = 3 u^{2} - 2 u - 1

因此

\frac{u ^{3} + 2 u ^{2} - 2 u - 1}{u - 1} = u^{2} + \frac{3 u ^{2} - 2 u - 1}{u - 1}

= u^{2} + \frac{3 u ^{2} - 2 u - 1}{u - 1}

对

\frac{3 u ^{2} - 2 u - 1}{u - 1}

做除法:

\frac{3 u ^{2} - 2 u - 1}{u - 1} = 3 u + \frac{u - 1}{u - 1}

将分子

3 u^{2} - 2 u - 1

与除数

u - 1

的首项系数相除：

\frac{3 u ^{2}}{u} = 3 u

商 = 3 u

将

u - 1

乘以

3 u : 3 u^{2} - 3 u

将

3 u^{2} - 2 u - 1

减去

3 u^{2} - 3 u

得到新的余数

余数 = u - 1

因此

\frac{3 u ^{2} - 2 u - 1}{u - 1} = 3 u + \frac{u - 1}{u - 1}

= u^{2} + 3 u + \frac{u - 1}{u - 1}

对

\frac{u - 1}{u - 1}

做除法:

\frac{u - 1}{u - 1} = 1

将分子

u - 1

与除数

u - 1

的首项系数相除：

\frac{u}{u} = 1

商 = 1

将

u - 1

乘以

1 : u - 1

将

u - 1

减去

u - 1

得到新的余数

余数 = 0

因此

\frac{u - 1}{u - 1} = 1

= u^{2} + 3 u + 1

= u^{2} + 3 u + 1

= (u - 1) (u^{2} + 3 u + 1)

= (u - 1) (u - 1) (u^{2} + 3 u + 1)

整理后得

= (u - 1)^{2} (u^{2} + 3 u + 1)

(u - 1)^{2} (u^{2} + 3 u + 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u - 1 = 0 or u^{2} + 3 u + 1 = 0

解

u - 1 = 0 : u = 1

u - 1 = 0

将

1

到右边

u - 1 = 0

两边加上

1

u - 1 + 1 = 0 + 1

化简

u = 1

u = 1

解

u^{2} + 3 u + 1 = 0 : u = \frac{- 3 + 5}{2}, u = \frac{- 3 - 5}{2}

u^{2} + 3 u + 1 = 0

使用求根公式求解

u^{2} + 3 u + 1 = 0

二次方程求根公式:

若

a = 1, b = 3, c = 1

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot 1 = 5

3^{2} - 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 3^{2} - 4

3^{2} = 9

= 9 - 4

数字相减:

9 - 4 = 5

= 5

u_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 3 + 5}{2 \cdot 1}, u_{2} = \frac{- 3 - 5}{2 \cdot 1}

u = \frac{- 3 + 5}{2 \cdot 1} : \frac{- 3 + 5}{2}

\frac{- 3 + 5}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 3 + 5}{2}

u = \frac{- 3 - 5}{2 \cdot 1} : \frac{- 3 - 5}{2}

\frac{- 3 - 5}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 3 - 5}{2}

二次方程组的解是:

u = \frac{- 3 + 5}{2}, u = \frac{- 3 - 5}{2}

解为

u = 1, u = \frac{- 3 + 5}{2}, u = \frac{- 3 - 5}{2}

u = 1, u = \frac{- 3 + 5}{2}, u = \frac{- 3 - 5}{2}

验证解

找到无定义的点（奇点）:

u = 0

取

- 4 + u + u^{2} + \frac{1}{u ^{2}} + \frac{1}{u}

的分母，令其等于零

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

u = 0

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = 1, u = \frac{- 3 + 5}{2}, u = \frac{- 3 - 5}{2}

u = cot (x)

代回

cot (x) = 1, cot (x) = \frac{- 3 + 5}{2}, cot (x) = \frac{- 3 - 5}{2}

cot (x) = 1, cot (x) = \frac{- 3 + 5}{2}, cot (x) = \frac{- 3 - 5}{2}

cot (x) = 1 : x = \frac{π}{4} + πn

cot (x) = 1

cot (x) = 1

的通解

cot (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

cot (x) = \frac{- 3 + 5}{2} : x = arccot (\frac{- 3 + 5}{2}) + πn

cot (x) = \frac{- 3 + 5}{2}

使用反三角函数性质

cot (x) = \frac{- 3 + 5}{2}

cot (x) = \frac{- 3 + 5}{2}

的通解

cot (x) = - a \Rightarrow x = arccot (- a) + πn

x = arccot (\frac{- 3 + 5}{2}) + πn

x = arccot (\frac{- 3 + 5}{2}) + πn

cot (x) = \frac{- 3 - 5}{2} : x = arccot (\frac{- 3 - 5}{2}) + πn

cot (x) = \frac{- 3 - 5}{2}

使用反三角函数性质

cot (x) = \frac{- 3 - 5}{2}

cot (x) = \frac{- 3 - 5}{2}

的通解

cot (x) = - a \Rightarrow x = arccot (- a) + πn

x = arccot (\frac{- 3 - 5}{2}) + πn

x = arccot (\frac{- 3 - 5}{2}) + πn

合并所有解

x = \frac{π}{4} + πn, x = arccot (\frac{- 3 + 5}{2}) + πn, x = arccot (\frac{- 3 - 5}{2}) + πn

以小数形式表示解

x = \frac{π}{4} + πn, x = 1.93566 \dots + πn, x = 2.77672 \dots + πn

作图

流行的例子

tan (a) = 0

cos^2(x)+3|cos(x)|-1=0

cos^{2} (x) + 3 ∣ cos (x) ∣ - 1 = 0

cos^5(x)=sin(75)

cos^{5} (x) = sin (7 5^{\circ})

csc^2(x)=sec(x)

csc^{2} (x) = sec (x)

(2cos(x)-sin^2(x))=1+cos^2(x)

(2 cos (x) - sin^{2} (x)) = 1 + cos^{2} (x)